The method for solving the KdV-equation

The method for solvin g the KdV-equation Dmitry Levko Abstract : T he method f or solvin g the KdV a re considere d. 23 May 2007 In [1] the simplest and direct method for cons truction of the solutions of nonlinear equations was considered. It is allow expressing the solutions of nonlinear equations of special class through the solutions of linear or nonlinear equations wi th known solutions. Analogous methods already are known for linear and nonlinear Schrödinger equations [2 ]-[3]. But method are presented in [1] contain some inaccuracies. It don’t allow step by step, without som e additional desires, to find the solution. In this work the improvement of method [1] are considered. For demonstration of the modified method the first example from [1] was chosen. Let us consider the linear equation 0 U U xxx t = + . (1) The new variable in this equation are t kx z ω + = , where k – wave number, ω – frequency. Then equation (1) we can rewrite in form: 0 U k U zzz 3 z = + ω . (2) The solution of such equation are considered in form ) z sin( A ) z ( U ⋅ = , (3) where A some constant. The dispersion correlation of (2) is next 3 k = ω . (4) From classical mechanics are know n that the or dinary differential equation of motion can be represent in form U ) U ( V U k U 3 zz ∂ ∂ = β + ω − = , (5) where ) U ( V – is the potential in which th e image point is m ove. From (3) and (2) we can find that 0 = β . Then we define the potential af ter the one integration. It is: λ + ω − = 2 3 U k 2 ) U ( V . (6) Here λ – some constant. Now let us consider the KdV 0 Q QQ 6 Q xxx x t = + + . (7) The new variable t kx z ω ′ − = . Analogous operations as for (1) can be made fo r (7). With new variable (7) are rewritten 0 Q k kQQ 6 Q zzz 3 z z = + + ω ′ − . (8) The potential of this equation has the form λ + β + − ω ′ = Q Q k 1 Q k 2 ) Q ( V 3 2 2 3 , (9) where β and λ – constants. From asymptotical equal ity to zero on inf inity the function ) z ( Q and it’s derivatives are got 0 = λ = β . To express the solution of (7) th rough the solution (1) let us assume that both solutions are re- lated by the correlation: ) Q ( g U = . (10) The function ) Q ( g can be finding from next expression [4] ) Q ( V ) g ( V ) Q ( g = ′ . (11) Then the system (1), (7) is given the equality ∫∫ − ω ′ = ω − λ 3 2 2 3 2 3 Q k 1 Q k 2 dQ g k 2 dg . (12) The solution of (12) in connect with (3) is dete rmined the functions g and Q. The first integral ω λ ω 3 3 k 2 g arcsin k 2 , and the second ω ′ − + ω ′ − − ω ′ kQ 2 1 1 kQ 2 1 1 ln k 2 3 . We can find the constant λ if we equate (3) to expression fo r function g that we got from (12) 2 A 2 = λ . The equality of phases for sin is allowed to determine the so lution of (7) ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ω ω ′ ⋅ ω ′ = 2 z h sec k 2 ) z ( Q 2 . To get the known soliton solution we need in la s t expression to suppose next dispersion corre- lation 3 k 4 = ω ′ . (13) Then finally ) t k 4 x ( k h sec k 2 ) z ( Q 2 2 2 − ⋅ = . (14) The given argumentations can be apply to all ex am ples in [1]. It’ll g ive the same results. Literature 1. D. Bazeia, A. Das, L. Losano and A. Silva. Arxiv: nlin.SI / 0703035v1 2. G. L Lamb, Jr. Elements of Solitons Theory. Mir, Moscow, 1983 (in Russian) 3. D. Levko. arXiv: nlin.SI / 0612027 4. D. Bazeia, L. Losano, and J.M.C. Malbouisson, Phys. Rev. D 66 , 101701(R) (2002) Dmitry Levko: Institute of Physics National Ukrainian Academy of Sciences d.le vko@gmail.com